BZOJ2120 数颜色

恩。。什么树状数组套主席树？大概是可以修改并且支持前缀减法的数据结构吧。。。

咦。可以离线？上莫队不就行了，干嘛要数据结构。。。

哦还要修改，那就三维莫队就好了，happy ending！

 

  1 /**************************************************************
  2     Problem: 2120
  3     User: rausen
  4     Language: C++
  5     Result: Accepted
  6     Time:488 ms
  7     Memory:5540 kb
  8 ****************************************************************/
  9  
 10 #include <cstdio>
 11 #include <cmath>
 12 #include <algorithm>
 13  
 14 using namespace std;
 15 const int N = 1e4 + 5;
 16 const int cnt_C = 1e6 + 5;
 17 const int Maxlen = N * 35;
 18  
 19 int n, m, sz, cnt_b;
 20 int w[N], c[N], pre[N];
 21 int now;
 22 int cnt_c, cnt_q;
 23 int ans[N], vis[N], cnt[cnt_C];
 24 char buf[Maxlen], *ch = buf;
 25 int Len;
 26  
 27 struct oper_change {
 28     int x, y, pre;
 29     oper_change() {}
 30     oper_change(int _x, int _y, int _p) : x(_x), y(_y), pre(_p) {}
 31 } oc[N];
 32  
 33 struct oper_query {
 34     int x, y, id, t;
 35     oper_query() {}
 36     oper_query(int _x, int _y, int _i, int _t) : x(_x), y(_y), id(_i), t(_t) {}
 37      
 38     inline bool operator < (const oper_query &X) const {
 39         return w[x] == w[X.x] ? w[y] == w[X.y] ? t < X.t : w[y] < w[X.y] : w[x] < w[X.x];
 40     }
 41 } oq[N];
 42  
 43 inline int read() {
 44     int x = 0;
 45     while (*ch < '0' || '9' < *ch) ++ch;
 46     while ('0' <= *ch && *ch <= '9')
 47         x = x * 10 + *ch - '0', ++ch;
 48     return x;
 49 }
 50  
 51 inline void reverse(int p) {
 52     if (vis[p]) {
 53         if ((--cnt[c[p]]) == 0) --now;
 54     } else {
 55         if ((++cnt[c[p]]) == 1) ++now;
 56     }
 57     vis[p] ^= 1;
 58 }
 59  
 60 inline int update(int p, int C) {
 61     if (vis[p]) {
 62         reverse(p);
 63         c[p] = C;
 64         reverse(p);
 65     } else c[p] = C;
 66 }
 67  
 68 int main() {
 69     Len = fread(ch, 1, Maxlen, stdin);
 70     buf[Len] = '\0';
 71     int i, j, l, r, x, y;
 72     char op;
 73     n = read(), m = read();
 74     sz = (int) pow(n, 2.0 / 3);
 75     if (!sz) sz = 1;
 76     for (i = 1; i <= n; ++i) {
 77         c[i] = pre[i] = read();
 78         if (i % sz == 1 || sz == 1) ++cnt_b;
 79         w[i] = cnt_b;
 80     }
 81     for (i = 1; i <= m; ++i) {
 82         while (*ch != 'Q' && *ch != 'R') ++ch;
 83         op = *ch;
 84         x = read(), y = read();
 85         if (op == 'R')
 86             oc[++cnt_c] = oper_change(x, y, pre[x]), pre[x] = y;
 87         else {
 88             if (x > y) swap(x, y);
 89             oq[++cnt_q] = oper_query(x, y, cnt_q, cnt_c);
 90         }
 91     }
 92     sort(oq + 1, oq + cnt_q + 1);
 93     for (i = l = 1, r = oq[0].t = 0; i <= cnt_q; ++i) {
 94         for (j = oq[i - 1].t + 1; j <= oq[i].t; ++j) update(oc[j].x, oc[j].y);
 95         for (j = oq[i - 1].t; j > oq[i].t; --j) update(oc[j].x, oc[j].pre);
 96         while (r < oq[i].y) reverse(++r);
 97         while (r > oq[i].y) reverse(r--);
 98         while (l > oq[i].x) reverse(--l);
 99         while (l < oq[i].x) reverse(l++);
100         ans[oq[i].id] = now;
101     }
102     for (i = 1; i <= cnt_q; ++i)
103         printf("%d\n", ans[i]);
104     return 0;
105 }
View Code

 

posted on 2015-03-09 20:53  Xs酱~  阅读(177)  评论(0)  编辑  收藏  举报